Mohr's circle

This article is part of the solid mechanics course, aimed at engineering students. Please leave feedback in the discussion section above.

Introduction

How would you break a metal rod if you could only use your hands?

1. Pull it apart or compress it (not usually the easiest way)

2. Twist it

3. Bend it

Each of these methods induce stress into the rod in a different way. Mohr's circle helps analyse this. For now, only the first two ways will be analysed.

1.Tension/compression:

Below is a diagram of a rod with a circular cross-section that is subjected to a tensile force at either end.(When you pull the rod apart you are exerting a tensile force). Let us look at a square element at the surface of the rod:

How do you imagine the square element would react to the tensile force?

The force will 'stretch' the rod and the square element, as shown below:

Therefore the force applied at either end has produced a tensile stress, σx{\displaystyle \sigma _{x}}, on the element, as shown below:

where: σx=FA{\displaystyle \sigma _{x}={\frac {F}{A}}}

Similarly, a Compressive force on the rod would induce a compressive stress on the square element, as shown below:

2. Torsion

In this case, the rod is subjected to a torque (twisting force) at either end:

How do you imagine the element would stretch this time?

Answer: The left side of the square element would stretch upwards, the right side would stretch downwards, as shown below. The stress induced by this shearing motion is called shear stress. It's symbol is, τ{\displaystyle \tau }. It is shown on the square element below:

Although it may be more difficult to visualize, there is also shear stress on the horizontal edges of the square element. The square element is drawn as follows:

If the torque was applied in the opposite direction, the shear stress on the element would look like this:

Take θ{\displaystyle \theta } to be 45 degrees. How do you imagine the element would react to the torque in this case? This one is more difficult to imagine. It would stretch as shown below:

Thus, you would draw the stresses on the element like so:

Note that there is no shear stress acting on the element at this orientation. When there is no shear stress acting on the element, the element is called the "principal element", and the 2 stresses on the element σ1{\displaystyle \sigma _{1}} and σ2{\displaystyle \sigma _{2}} are known as the principal stresses.

The objective of the Mohr's circle method is to find the orientation of the principal element (i.e.θ{\displaystyle \theta }, which for this simple case was 45 degrees), and find the values of σ1{\displaystyle \sigma _{1}} and σ2{\displaystyle \sigma _{2}}.

We take the shear stress as negative, because the shear stress at surface A tries to rotate the square element in an anticlockwise direction (about the centre of the element). This is the general convention used.

Finally, to transform into the coordinate system of the principal axes, rotate the original square element by θ{\displaystyle \theta } degrees clockwise (because you draw 2θ{\displaystyle \theta } as a clockwise angle in the above diagram):

A rod is subjected to a tensile force and a torque, as shown below. Use Mohr's circle to work out the principal stresses and draw the rotated square element. (I recommend you try this first before seeing the answer)

Notice that 66MPa is drawn as a tensile stress (as σ1{\displaystyle \sigma _{1}\,} is positive), and 40MPa as a compressive stress (as σ2{\displaystyle \sigma _{2}\,} is negative)

That's it! If you have found this article useful, please comment in the discussion section (at the top of the page), as this will help me decide whether to write more articles like this. Also please comment if there are other topics you want covered, or you would like something in this article to be written in more detail.